On a Cardinality-Constrained Transportation Problem With
Market Choice
Matthias Walter∗1 , Pelin Damcı-Kurt†2 , Santanu S. Dey‡3 , and Simge Küçükyavuz§2
1
Institut für Mathematische Optimierung, Otto-von-Guericke-Universität Magdeburg,
Magdeburg, Germany
2
Department of Integrated Systems Engineering, The Ohio State University, Columbus, OH
43210, United States
3
School of Industrial and Systems Engineering, Georgia Institute of Technology, Atlanta,
GA 30332, United States
June 13, 2015
Abstract
It is well-known that the intersection of the matching polytope with a cardinality constraint is integral [8]. In this note, we prove a similar result for the polytope corresponding to the transportation
problem with market choice (TPMC) (introduced in [4]) when the demands are in the set {1, 2}. This
result generalizes the result regarding the matching polytope. The result in this note uses the fact that
some special classes of minimum weight perfect matching problem with a cardinality constraint on a
subset of edges can be solved in polynomial time.
1
Introduction and Main Result
1.1
Transportation Problem with Market Choice
The transportation problem with market choice (TPMC), introduced in the paper [4], is a transportation
problem in which suppliers with limited capacities have a choice of which demands (markets) to satisfy. If a
market is selected, then its demand must be satisfied fully through shipments from the suppliers. If a market
is rejected, then the corresponding potential revenue is lost. The objective is to minimize the total cost of
shipping and lost revenues. See [5, 7, 9] for approximation algorithms and heuristics for several other supply
chain planning and logistics problems with market choice.
Formally, we are given a set of supply and demand nodes that form a bipartite graph G = (V1 ∪ V2 , E).
The nodes in set V1 represent the supply nodes, where for i ∈ V1 , si ∈ N represents the capacity of supplier
i. The nodes in set V2 represent the potential markets, where for j ∈ V2 , dj ∈ N represents the demand of
market j. The edges between supply and demand nodes have weights that represent shipping costs we , where
e ∈ E. For each j ∈ V2 , rj is the revenue lost if the market j is rejected. Let x{i,j} be the amount of demand
of market j satisfied by supplier i for {i, j} ∈ E, and let zj be an indicator variable taking a value 1 if market
∗ [email protected]
† [email protected]
‡ [email protected]
§ [email protected]
1
j is rejected and 0 otherwise. A mixed-integer programming (MIP) formulation of the problem is given where
the objective is to minimize the transportation costs and the lost revenues due to unchosen markets:
X
min
|E|
x∈R+ , z∈{0,1}|V2 |
s.t.
we xe +
e∈E
X
X
rj zj
(1)
j∈V2
x{i,j} = dj (1 − zj )
∀j ∈ V2
(2)
x{i,j} ≤ si
∀i ∈ V1 .
(3)
i:{i,j}∈E
X
j:{i,j}∈E
We refer to the formulation (1)-(3) as TPMC. The first set of constraints (2) ensures that if market j ∈ V2
is selected (i.e., zj = 0), then its demand must be fully satisfied. The second set of constraints (3) model the
supply restrictions.
TPMC is strongly NP-complete in general [4]. [1] give polynomial-time reductions from this problem to
the capacitated facility location problem [6], thereby establishing approximation algorithms with constant
factors for the metric case and a logarithmic factor for the general case.
1.2
TMPC with dj ∈ {1, 2} for all j ∈ V2 and the Matching Polytope
When dj ∈ {1, 2} for each demand node j ∈ V2 , TPMC is polynomially solvable [4]. This is proven through a
reduction to a minimum weight perfect matching problem on a general (non-bipartite) graph G0 = (V 0 , E 0 );
see [4]. We call this special class of the problem, the simple TPMC problem in the rest of this note.
Observation 1 (Simple TPMC generalizes Matching on General Graphs). The matching problem can be
seen as a special case of the simple TPMC problem. Let G = (V, E) be a graph with n vertices and m edges.
We construct a bipartite graph Ĝ = (V̂ 1 ∪ V̂ 2 , Ê) as follows: V̂ 1 is a set of n vertices corresponding to the n
vertices in G, and V̂ 2 corresponds to the set of edges of G, i.e., V̂ 2 contains m vertices. We use {i, j} to refer
to the vertex in V̂ 2 corresponding to the edge {i, j} in E. The set of edges in Ê are of the form {i, {i, j}} and
{j, {i, j}} for every i, j ∈ V such that {i, j} ∈ E. Now we can construct (the feasible region of ) an instance
of TPMC with respect to Ĝ = (V̂ 1 ∪ V̂ 2 , Ê) as follows:
m
2
Q = {(x, z) ∈ R2m
+ × R | x{i,e} + x{j,e} +2ze = 2 ∀e = {i, j} ∈ V̂
X
x{i,{i,j}} ≤ 1 ∀i ∈ V̂ 1
(4)
(5)
j:{i,j}∈E
ze ∈ {0, 1} ∀e ∈ V̂ 2 }.
(6)
Clearly there is a bijection between the set of matchings in G and the set of solutions in Q. Moreover, let
H := {(x, z, y) ∈ R2m × Rm × Rm | (x, z) ∈ Q, y = e − z},
where e is the all ones vector in Rm . Then we have that the incidence vector of all the matchings in G = (V, E)
is precisely the set projy (H).
Note that the instances of the form of (4)-(6) are special cases of simple TPMC instances, since in these
instances all si ’s are restricted to be exactly 1 and all dj ’s are restricted to be exactly 2.
1.3
Simple TPMC with Cardinality Constraint: Main Result
An important and natural constraint that one may add to the TPMC problem is that of a service level,
that is the number of rejected
P markets is restricted to be at most k. This restriction can be modelled using
a cardinality constraint,
j∈V2 zj ≤ k, appended to (1)-(3). We call the resulting problem cardinalityconstrained TPMC (CCTPMC). If we are able to solve CCTPMC in polynomial-time, then we can solve
2
TPMC in polynomial time by solving CCTPMC for all k ∈ {0, . . . , |V2 |}. Since TPMC is NP-hard, CCTPMC
is NP-hard in general.
In this note, we examine the effect of appending a cardinality constraint to the simple TPMC problem.
Theorem 1. Given an instance of TPMC with V2 , the set of demand nodes, and E, the set of edges, let
|E|
X ⊆ R+ × {0, 1}|V2 | be the set of feasible solutions of this instance of TPMC. Let k ∈ Z+ and k ≤ |V2 |.
P
|E|
k
Let X k := conv(X ∩ {(x, z) ∈ R+ × {0, 1}|V2 | |
j∈V2 zj ≤ k}). If dj ≤ 2 for all j ∈ V2 , then X =
P
|E|
conv(X) ∩ {(x, z) ∈ R+ × [0, 1]|V2 | | j∈V2 zj ≤ k}.
Our proof of Theorem 1 is presented in Section 2. We note that the result of Theorem 1 holds even when
P
|E|
|E|
k
|V2 |
X
| j∈V2 zj ≥ k}) or conv(X ∩ {(x, z) ∈ R+ × {0, 1}|V2 | |
P is defined as conv(X ∩ {(x, z) ∈ R+ × {0, 1}
j∈V2 zj = k}).
By invoking the ellipsoid algorithm and the use of Theorem 1 we obtain the following corollary.
Corollary 1. Cardinality constrained simple TPMC is polynomially solvable.
We note that, as a consequence of Theorem 1 (but also inherent in our proof), a special class of minimum
weight perfect matching problem with a cardinality constraint on a subset of edges can be solved in polynomial
time: Simple TPMC can be reduced to a minimum weight perfect matching problem on a general (nonbipartite) graph G0 = (V 0 , E 0 ) [4]. Therefore, it is possible to reduce CCTPMC with dj ≤ 2 for all j ∈ V2
to a minimum weight perfect matching problem with a cardinality constraint on a subset of edges. Hence,
Corollary 1 implies that a special class of minimum weight perfect matching problems with a cardinality
constraint on a subset of edges can be solved in polynomial time.
Note that the intersection of the perfect matching polytope with a cardinality constraint on a strict subset
of edges is not always integral.
Example 1. Consider the cycle C4 of length 4 with edge set E = {{1, 2}, {2, 3}, {3, 4}, {1, 4}}, and the
cardinality constraint x12 + x34 = 1. The only perfect matchings are {{1, 2}, {3, 4}} and {{1, 4}, {2, 3}} for
which the cardinality constraint has activity 2 and 0, respectively. Thus the perfect matching polytope is a
line which is intersected by the hyperplane defined by the cardinality constraint in the (fractional) center.
To the best of our knowledge, the complexity status of minimum weight perfect matching problem on a
general graph with a cardinality constraint on a subset of edges is open. This can be seen by observing that
if one can solve minimum weight perfect matching problem with a cardinality constraint on a subset of edges
in polynomial time, then one can solve the exact perfect matching problem in polynomial time; see discussion
in the last section in [2].
Finally we ask the natural question: Does the statement of Theorem 1 hold when dj ≤ 2 does not hold
for every j? The next example illustrates that the statement does not hold in such case.
Example 2. Consider an instance of TPMC where G = (V1 ∪ V2 , E) is a bipartite graph with
V1 = {i1 , i2 , i3 , i4 , i5 , i6 },
V2 = {j1 , j2 , j3 , j4 },
E = {{i1 , j1 }, {i2 , j2 }, {i3 , j3 }, {i4 , j1 }, {i4 , j4 }, {i5 , j2 }, {i5 , j4 }, {i6 , j3 }, {i6 , j4 }},
si = 1, i ∈ V1 ,
dj1 = dj2 = dj3 = 2, dj4 = 3.
p
n
For
Pn k = 2 it can be verified that we obtain a non-integer extreme point of conv(X) ∩ {(x, z) ∈ R+ × [0, 1] |
j=1 zj ≤ k}, given by x{i1 ,j1 } = x{i2 ,j2 } = x{i3 ,j3 } = x{i4 ,j1 } = x{i4 ,j4 } = x{i5 ,j2 } = x{i5 ,j4 } = x{i6 ,j3 } =
x{i6 ,j4 } = z1 = z2 = z3 = z4 = 21 . To see this, consider the face defined by the supply constraints of nodes
{i4 , i5 , i6 } and observe that this face has precisely two solutions
having 1 and 3 markets, respectively.
Pn
Therefore, X k 6= conv(X) ∩ {(x, z) ∈ Rp+ × [0, 1]n | j=1 zj ≤ k} in this example.
3
sa = 2
sb = 1
sc = 1
a1
a
a2
x
dx = 1
y
dy = 1
x
x̂
y
ŷ
b1
b
c1
c
z
dz = 2
z1
d1
sd = 3
d2
d
z2
d3
Figure 1: Improved Reduction to a Matching Problem
2
Proof of Theorem 1
To prove Theorem 1 we use an improved reduction to a minimum weight matching problem (compared to
the reduction in [4]) and then use the well-known adjacency properties of the vertices of the perfect matching
polytope. Since the integrality result does not hold for the perfect matching polytope on a general graph with
a cardinality constraint on any subset of edges, as illustrated in Example 1, we need to refine the adjacency
criterion.
We begin with some notation. For a graph G = (V, E) with node set V and edge set E, and a node v ∈ V ,
we denote by δ(v) := δG (v) := {e ∈ E | v ∈ e} thePset of edges incident to v. For a vector x ∈ R|E| and a
subset F ⊆ E of its ground set, we define x(F ) := f ∈F xf .
We now describe the improved reduction to a minimum weight matching problem. Consider a simple
TPMC instance on a graph G = (V1 ∪ V2 , E) with supplies s ∈ N|V1 | , demands d ∈ {1, 2}|V2 | , edge weights
w ∈ R|E| , and revenues r ∈ R|V2 | . Let Dk = {j ∈ V2 | dj = k} be the partitioning of V2 into two classes
corresponding to the demands.
b 1 ∪D1 ∪D2 and edges E1 ∪E2 ∪F1 ∪F2
We create the auxiliary graph G∗ (see Figure 1) with nodes V1s ∪D1 ∪D
2
2
with
V1s = {i` | i ∈ V1 and ` ∈ {1, 2, . . . , si }},
b 1 = {ĵ | j ∈ D1 },
D
D2k = {jk | j ∈ D2 } for k = 1, 2,
E1 = {{i` , j} | {i, j} ∈ E, ` ∈ {1, 2, . . . , si } and j ∈ D1 },
E2 = {{i` , jk } | {i, j} ∈ E, ` ∈ {1, 2, . . . , si }, j ∈ D2 and k ∈ {1, 2}},
F1 = {{j, ĵ} | j ∈ D1 }, and
F2 = {{j1 , j2 } | j ∈ D2 }.
In the construction every node i ∈ V1 with supply si is split into si identical nodes with intended supply
value of 1. Furthermore, to every node j ∈ V2 with demand 1 we attach an edge with a dead end ĵ, and every
node j ∈ V2 with demand 2 is split into nodes j1 and j2 which are connected by an edge. Note that this is a
polynomial construction, because the supply, si , is at most 2|V2 | for any i ∈ V1 .
|E|
Lemma 1. Let X ⊆ R+ × {0, 1}|V2 | be the set of feasible solutions of a simple TPMC instance on a graph
G = (V1 ∪ V2 , E) with supplies s ∈ N|V1 | and demands d ∈ {1, 2}|V2 | . Let the sets Dk and the auxiliary graph
G∗ be defined as above.
4
Then P := conv(X) is equal to the projection of the face of the matching polytope Pmatch (G∗ ) of G∗
Q := {y ∈ Pmatch (G∗ ) | y(δ(v)) = 1 for all v ∈ D1 ∪ D21 ∪ D22 }
Psi
Psi
(y{i` ,j1 } + y{i` ,j2 } ) for
y{i` ,j} for {i, j} ∈ E and j ∈ D1 , x{i,j} = `=1
via the map π defined by x{i,j} = `=1
{i, j} ∈ E and j ∈ D2 , zj = y{j,ĵ} for j ∈ D1 , and zj = y{j1 ,j2 } for j ∈ D2 .
Proof. We first show π(Q) ⊆ P . Let y be a vertexPof Q and (x, z) = π(y) be the projection.
si
∗
Clearly, for all i ∈ V1 we have x(δG (i)) =
`=1 y(δG (i` )) ≤ si , i.e., (x, z) satisfies (3). For every
node j ∈ D1 we have x(δG (j)) + zj = y(δG∗ (j) \ {{j, ĵ}}) + y{j,ĵ} = y(δG∗ (j)) = 1. Furthermore, for
every node j ∈ D2 we have x(δG (j)) + 2zj = y(δG∗ (j1 ) \ {{j1 , j2 }}) + y(δG∗ (j1 ) \ {{j1 , j2 }}) + 2y{j1 ,j2 } =
y(δG∗ (j1 )) + y(δG∗ (j2 )) = 2. Hence, (x, z) satisfies (2) proving (x, z) ∈ conv(X) since z is binary.
We now show P ⊆ π(Q) for which it suffices to consider only integer points in P since both polytopes are
integral. Note that P is integral since for integral z the remaining system is totally unimodular with integral
|E|
right-hand side. Let (x, z) ∈ P ∩ (Z+ × {0, 1}|V2 | ) be an integral point in P . For j ∈ D1 , let ej ∈ E be
the unique edge with x{i,j} > 0, and for j ∈ D2 , let {ej , fj } be the set of edges incident to j with positive
x-value. Observe that if ej = fj holds, then xej = 2, and otherwise xej = xfj = 1.
Construct a matching M satisfying
M = {{j, ĵ} | j ∈ D1 with zj = 1} ∪ {{j1 , j2 } | j ∈ D2 with zj = 1}
(7)
∪ {{i` , j} | j ∈ D1 and i ∈ ej with zj = 0}
(8)
∪ {{i` , j1 } | j ∈ D2 and i ∈ ej with zj = 0}
(9)
∪ {{i` , j2 } | j ∈ D2 and i ∈ fj with zj = 0}
(10)
choosing ` in (8)–(10) such that every node i` ∈ V1s has at most one incident edge in M . This is possible
since for each i ∈ V1 , G∗ has si identical copies i1 , . . . , isi and M has to contain at most x(δG (i)) ≤ si edges
incident to one of the copies because x is integral.
We first prove that M is indeed a matching. A node j ∈ D1 is matched either to ĵ (if zj = 1) or by ej .
Similarly, either j1 and j2 are matched by the edge {j1 , j2 } (again if zj = 1) or by ej and fj , respectively.
The fact that M projects to (x, z) is easy to check by the construction of M according to (7)–(10). This
concludes the proof.
We now turn to the proof of Theorem 1. By definition of the projection map π in Lemma 1, the equation
z(V2 ) = k corresponds to the equation y(F1 ∪ F2 ) = k in Q, that is,
|E|
P ∩ {(x, z) ∈ R+ × [0, 1]|V2 | | z(V2 ) = k} = {π(y) | y ∈ Q with y(F1 ∪ F2 ) = k}
holds. Hence, in order to show that the former is integral (and since π projects integral vectors to integral
vectors), it suffices to prove the following claim:
|E|
Claim 1. Let X ⊆ R+ × {0, 1}|V2 | be the set of feasible solutions of a simple TPMC instance on a graph
G = (V1 ∪ V2 , E) with supplies s ∈ N|V1 | and demands d ∈ {1, 2}|V2 | . Let the sets Dk and the auxiliary graph
G∗ be defined as above and let Q be as in Lemma 1.
Then {y ∈ Q | y(F1 ∪ F2 ) = k} is an integral polytope for any integer k ∈ Z+ .
Proof. Let H = {y | y(F1 ∪ F2 ) = k} denote the intersecting hyperplane and assume, for the sake of
contradiction, that Q ∩ H is not integral. Then there must exist two adjacent (in Q) matchings M1 and
M2 defining an edge of Q that is intersected by H in its relative interior, i.e., |M1 ∩ (F1 ∪ F2 )| < k and
|M2 ∩ (F1 ∪ F2 )| > k.
By the adjacency characterization of the matching polytope [3], the symmetric difference C := M1 ∆M2
must be a connected component (a cycle or a path) in G∗ containing edges of M1 and M2 in an alternating
fashion.
5
We now verify that there must exist a path e-P -f in C of odd length consisting of two edges e, f ∈
C ∩ (F1 ∪ F2 ) and a subpath P in C \ (F1 ∪ F2 ): If for every choice of e, f ∈ M2 ∩ C ∩ (F1 ∪ F2 ) there exists
an edge belonging to M1 ∩ (F1 ∪ F2 ) in all subpath(s) e-P -f of C, then M2 can have at most one more edge
of F1 ∪ F2 than M1 in C. However since M2 contains at least two more edges of F1 ∪ F2 than M1 does, we
have that there exists a path e-P -f in C consisting of two edges e, f ∈ M2 ∩ C ∩ (F1 ∪ F2 ) and a subpath P
in C \ (F1 ∪ F2 ). Now since e-P -f is subpath of C and e, f ∈ M2 , we have that P is of odd length.
Clearly, since we have P ∩ (F1 ∪ F2 ) = ∅, all of P ’s edges must go between V1s and D1 ∪ (D21 ∪ D22 ). Since
P also has odd length, one of its endpoints is in V1s . But no edge in F1 ∪ F2 is incident to any node in V1s
which yields a contradiction.
Acknowledgements.
Pelin Damcı-Kurt and Simge Küçükyavuz are supported, in part, by NSF-CMMI grant 1055668. Santanu S.
Dey gratefully acknowledges the support of the Air Force Office of Scientific Research grant FA9550-12-1-0154.
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